Difference between revisions of "2010 AIME I Problems/Problem 6"
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== Problem == | == Problem == | ||
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>. | Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>. | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
<center><asy> | <center><asy> | ||
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); | import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); | ||
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Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>. | Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>. | ||
− | == Solution 2 == | + | === Solution 2 === |
− | It can be seen that the function <math> | + | It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations: |
− | <math> | + | <center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center> |
− | Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math> | + | Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>. |
− | == See | + | === Solution 3 === |
+ | |||
+ | Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that equality holds when <math>y = 1</math> and therefore when <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(1) = 1</math>. | ||
+ | |||
+ | Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality: | ||
+ | |||
+ | <math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math> | ||
+ | |||
+ | <math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus | ||
+ | |||
+ | <math>x - 2 \le Q'(x) \le 2x - 3</math> | ||
+ | |||
+ | For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However, | ||
+ | |||
+ | <math>\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1</math>, | ||
+ | |||
+ | and thus by the [[sandwich theorem]] <math>\lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | Let <math>Q(x) = P(x) - (x^2-2x+2)</math>, then <math>0\le Q(x) \le (x-1)^2</math> (note this is derived from the given inequality chain). Therefore, <math>0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2</math> for some real value A. | ||
+ | |||
+ | <math>Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}</math>. | ||
+ | |||
+ | <math>Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}</math> | ||
+ | |||
+ | === Solution 5 === | ||
+ | |||
+ | Let <math>P(x) = ax^2 + bx + c</math>. Plugging in <math>x = 1</math> to the expressions on both sides of the inequality, we see that <math>a + b + c = 1</math>. We see from the problem statement that <math>121a + 11b + c = 181</math>. Since we know the vertex of <math>P(x)</math> lies at <math>x = 1</math>, by symmetry we get <math>81a -9b + c = 181</math> as well. Since we now have three equations, we can solve this trivial system and get our answer of <math>\boxed{406}</math>. | ||
+ | |||
+ | === Solution 6 === | ||
+ | |||
+ | Similar to Solution 5, let <math>P(x) = ax^2 + bx + c</math>. Note that <math>(1,1)</math> is a vertex of the polynomial. Additionally, this means that <math>b = -2a</math> (since <math>\frac{-b}{2a}</math> is the minimum <math>x</math> point). Thus, we have <math>P(x) = ax^2 - 2ax + c</math>. Therefore <math>a - 2a + c = 1</math>. Moreover, <math>99a + c = 181</math>. And so our polynomial is <math>\frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}</math>. Plug in <math>x = 16</math> to get <math>\boxed{406}</math>. | ||
+ | |||
+ | == See Also == | ||
{{AIME box|year=2010|num-b=5|num-a=7|n=I}} | {{AIME box|year=2010|num-b=5|num-a=7|n=I}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:20, 17 December 2021
Contents
Problem
Let be a quadratic polynomial with real coefficients satisfying for all real numbers , and suppose . Find .
Solution
Solution 1
Let , . Completing the square, we have , and , so it follows that for all (by the Trivial Inequality).
Also, , so , and obtains its minimum at the point . Then must be of the form for some constant ; substituting yields . Finally, .
Solution 2
It can be seen that the function must be in the form for some real and . This is because the derivative of is , and a global minimum occurs only at (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at ). Substituting and we obtain two equations:
Solving, we get and , so . Therefore, .
Solution 3
Let ; note that . Setting , we find that equality holds when and therefore when ; this is true iff , so .
Let ; clearly , so we can write , where is some linear function. Plug into the given inequality:
, and thus
For all ; note that the inequality signs are flipped if , and that the division is invalid for . However,
,
and thus by the sandwich theorem ; by the definition of a continuous function, . Also, , so ; plugging in and solving, . Thus , and so .
Solution 4
Let , then (note this is derived from the given inequality chain). Therefore, for some real value A.
.
Solution 5
Let . Plugging in to the expressions on both sides of the inequality, we see that . We see from the problem statement that . Since we know the vertex of lies at , by symmetry we get as well. Since we now have three equations, we can solve this trivial system and get our answer of .
Solution 6
Similar to Solution 5, let . Note that is a vertex of the polynomial. Additionally, this means that (since is the minimum point). Thus, we have . Therefore . Moreover, . And so our polynomial is . Plug in to get .
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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